遍历一个字符串并用haskell中的子字符串替换单个字符 [英] Iterating through a String and replacing single chars with substrings in haskell

问题描述

我试图学习一些Haskell，我发现它很困难。我在使用
当前项目时遇到了一些问题。这个想法是，我必须通过一个字符串并用新的子字符串替换某些字符
。例如，如果我有一个字符串FLXF，我想用一个名为FLF的子字符串替换每个F
，结果应该是FLFLXFLF。现在我一直在研究这个
特定的问题几个小时。我一直在阅读类型，可能派上用场的不同功能（地图，折叠等），但我一直无法解决这个问题。

下面的代码是我有过的一些不同的尝试：

  apply :: String  - >字符串
 apply [] = [] 
 apply（x：xs）= if（x =='F'）
然后显示Hello
 apply xs 
 else（apply（xs））
  

这个例子在这里我只是试图展示你好，遇到'F'，但它显示的是，所以这显然不起作用。我真的不确定是否有其他陈述是这里的方法。我也在想功能图可能会做到这一点。这里我想到的代码可能看起来像这样：

  map（\xy  - > if y ==' F'然后是FLDelse y）FLF
  

但是这给我一个类型错误。所以你可以看到我迷路了。请原谅我对Haskell的可怜知识，但我仍然对它不熟悉。我真的希望你们中的一些人能够帮助我，或者让我朝着正确的方向前进。如果我对某些事情不清楚，请随时提问。

预先感谢您！

John

解决方案

  map（\xy  - > if y =='F'then FLDelse y）FLF
  

这几乎是正确的。

第一个...为什么函数需要两个参数？

  map（\y  - > ;如果y =='F'，那么FLDelse y）FLF
  

其余类型错误是因为然后分支给出了 String ，但是 else 分支给出了一个 Char （两个分支都必须给出相同类型的值）。因此，我们将使 else 分支给出 String 来代替（回想一下 String 是 [Char] ）的同义词：



  map（\y  - > if y =='F'thenFLDelse [y]）FLF
  

现在问题是这给你一个 [String] 值而不是 String 。所以我们将所有这些字符串连接在一起：



  concat（map（\y  - > if y =='F '然后是FLDelse [y]）FLF）
  

concat 和 map 已经足够普遍，以至于有一个标准函数可以将它们结合在一起。

  concatMap（\y  - > if y =='F'thenFLDelse [y]）FLF
  




I am trying to learn some Haskell and I find it difficult. I am having some issues with my current project. The idea is that I have to go through a String and substitute certain chars with new substrings. For instance if I have a String "FLXF" and I want to replace every F with a substring called "FLF" the result should be "FLFLXFLF". Now I have been working on this specific problem for hours. I have been reading up on types, different functions that might come in handy (map, fold, etc) and yet I have not been able to solve this problem.

The code below is some of the different tries I have had:

apply :: String -> String
apply []     = []
apply (x:xs) = if (x == 'F')
               then do show "Hello"
                       apply xs
               else (apply (xs))

This example here I was just trying to show hello every time I encountered a 'F', but all it shows is "", so this clearly does not work. I am really not sure an if else statement is the way to go here. I was also thinking the function map might do the trick. Here the code I was thinking about could look something like this:

map (\x y -> if y == 'F' then "FLD" else y) "FLF"

but that gives me a type error. So as you can see I am lost. Excuse me my poor knowledge to Haskell, but I am still new to it. I really hope some of you can help me out here or give me a push in the right direction. Feel free to ask questions if I have been unclear about something.

Thank you in advance!

John

解决方案

map (\x y -> if y == 'F' then "FLD" else y) "FLF"

This is nearly right.

First... why does the function take two arguments?

map (\y -> if y == 'F' then "FLD" else y) "FLF"

The remaining type error is because the then branch gives a String, but the else branch gives a Char (the two branches must each give a value of the same type). So we'll make the else branch give a String instead (recall that String is a synonym for [Char]):

map (\y -> if y == 'F' then "FLD" else [y]) "FLF"

Now the problem is that this gives you a [String] value instead of a String. So we'll concatenate all those strings together:

concat (map (\y -> if y == 'F' then "FLD" else [y]) "FLF")

This combination of concat and map is common enough that there's a standard function that combines them.

concatMap (\y -> if y == 'F' then "FLD" else [y]) "FLF"

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